If acoin roll at 1.5m/s and later rolls at 4.5m/s with an acceleration of 2 meters per squared second how long will it take to come to a stop, how far will it travel?
To find out how long it will take for the coin to come to a stop and the distance it will travel, we can use the equations of motion. Specifically, we can use the equation of motion for an object with constant acceleration:
v^2 = u^2 + 2as
where:
- v is the final velocity when the object comes to a stop (0 m/s in this case),
- u is the initial velocity (1.5 m/s),
- a is the acceleration (-2 m/s^2),
- and s is the distance traveled.
First, let's find the time it takes for the coin to come to a stop.
Given:
u = 1.5 m/s
v = 0 m/s
a = -2 m/s^2 (negative because it is decelerating)
Using the equation v = u + at, we can solve for t:
0 = 1.5 + (-2)t
-1.5 = -2t
t = -1.5 / -2
t ≈ 0.75 seconds
Therefore, it will take approximately 0.75 seconds for the coin to come to a stop.
Next, let's determine the distance traveled by the coin.
Using the equation v^2 = u^2 + 2as, we can solve for s:
0^2 = 1.5^2 + 2(-2)s
0 = 2.25 - 4s
4s = 2.25
s = 2.25 / 4
s ≈ 0.5625 meters
Therefore, the coin will travel approximately 0.5625 meters before coming to a stop.
To summarize:
- The coin will take approximately 0.75 seconds to come to a stop.
- It will travel approximately 0.5625 meters before stopping.